What Are You Talking About Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others) Total Submission(s): 15966    Accepted Submission(s): 5177 Problem Description Ignatius is so lucky that he met a Martian yesterday. But…

What Are You Talking About Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others) Total Submission(s): 16042    Accepted Submission(s): 5198 Problem Description Ignatius is so lucky that he met a Martian yesterday. But…

题意:单词翻译 思路:map #include<iostream> #include<stdio.h> #include<string.h> #include<map> using namespace std; map<string,string>mp; int main(){ string s1,s2; char c; cin>>s1; while(cin>>s1){ if(s1=="END")bre…

What Are You Talking About Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others) Total Submission(s): 20680    Accepted Submission(s): 6852 Problem Description Ignatius is so lucky that he met a Martian yesterday. But…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1052 Problem Description Here is a famous story in Chinese history. "That was about 2300 years ago. General Tian Ji was a high official in the country Qi. He likes to play horse racing with the king and…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 49436    Accepted Submission(s): 22871 Problem Description Nowadays, a kind of chess game called "Super Jumping!…

<题目链接> 题目大意: 意思是,从第1个字母到第2字母组成的字符串可由某一周期性的字串(“a”) 的两次组成,也就是aa有两个a组成: 第三行自然就是aabaab可有两个aab组成: 第四行aabaabaab可由三个aab组成: 第五行aabaabaabaab可有四个aab组成 解题分析: 求字符串的前缀是否为周期串,若是,打印循环节的长度及循环次数: #include <cstdio> #include <cstring> ; char s[M]; int nxt[…

Description John is the only priest in his town. October 26th is the John's busiest day in a year because there is an old legend in the town that the couple who get married on that day will be forever blessed by the God of Love. This year N couples p…

Parentheses Matrix Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0Special Judge Problem Description A parentheses matrix is a matrix where every element is eithe…

题意: 给你一个1e9-1e14的质数P,让你找出这个质数的前一个质数Q,然后计算Q!mod P 题解: 1e14的数据范围pass掉一切素数筛法,考虑Miller-Rabin算法. 米勒拉宾算法是一种判断素数的随机化算法,由于其随机性,它不能保证总是正确的,但其对于一个素数,总会返回素数的结果,对于一个合数,才有极小概率返回素数的结果(假阳性). 米勒拉宾算法对于单个素数的判断时间复杂度为$O(log^3n)$.(因为1e14相乘会爆longlong,模乘要写成龟速乘,因此要多一个log) 1…